What is the general solution for sec^2(x)+5tan(x)=7 ?

The general solution for sec^2(x)+5tan(x)=7 is x= pi/4+pin,x=-1.40564…+pin

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sec^2(x)+5tan(x)=7

Solution

sec^{2} (x) + 5 tan (x) = 7

Solution

x = \frac{π}{4} + πn, x = - 1.40564 \dots + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 80.53767 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sec^{2} (x) + 5 tan (x) = 7

Subtract

7

from both sides

sec^{2} (x) + 5 tan (x) - 7 = 0

Rewrite using trig identities

- 7 + sec^{2} (x) + 5 tan (x)

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

= - 7 + tan^{2} (x) + 1 + 5 tan (x)

Simplify

- 7 + tan^{2} (x) + 1 + 5 tan (x) : tan^{2} (x) + 5 tan (x) - 6

- 7 + tan^{2} (x) + 1 + 5 tan (x)

Group like terms

= tan^{2} (x) + 5 tan (x) - 7 + 1

Add/Subtract the numbers:

- 7 + 1 = - 6

= tan^{2} (x) + 5 tan (x) - 6

= tan^{2} (x) + 5 tan (x) - 6

- 6 + tan^{2} (x) + 5 tan (x) = 0

Solve by substitution

- 6 + tan^{2} (x) + 5 tan (x) = 0

Let:

tan (x) = u

- 6 + u^{2} + 5 u = 0

- 6 + u^{2} + 5 u = 0 : u = 1, u = - 6

- 6 + u^{2} + 5 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + 5 u - 6 = 0

Solve with the quadratic formula

u^{2} + 5 u - 6 = 0

Quadratic Equation Formula:

For

a = 1, b = 5, c = - 6

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 6) = 7

5^{2} - 4 \cdot 1 \cdot (- 6)

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 7}{2 \cdot 1}, u_{2} = \frac{- 5 - 7}{2 \cdot 1}

u = \frac{- 5 + 7}{2 \cdot 1} : 1

\frac{- 5 + 7}{2 \cdot 1}

Add/Subtract the numbers:

- 5 + 7 = 2

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 5 - 7}{2 \cdot 1} : - 6

\frac{- 5 - 7}{2 \cdot 1}

Subtract the numbers:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 12}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{12}{2}

Divide the numbers:

\frac{12}{2} = 6

= - 6

The solutions to the quadratic equation are:

u = 1, u = - 6

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - 6

tan (x) = 1, tan (x) = - 6

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 6 : x = arctan (- 6) + πn

tan (x) = - 6

Apply trig inverse properties

tan (x) = - 6

General solutions for

tan (x) = - 6

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- 6) + πn

x = arctan (- 6) + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = arctan (- 6) + πn

Show solutions in decimal form

x = \frac{π}{4} + πn, x = - 1.40564 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)+5tan(x)=7 ?
The general solution for sec^2(x)+5tan(x)=7 is x= pi/4+pin,x=-1.40564…+pin